242MidTermSol2005

242MidTermSol2005 - Question 1 Part a 0418 10 12 1 12 10 =...

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Unformatted text preview: Question 1 Part a) % 0418 . 10 ) 12 / 1 ( 12 10 . = ⇒ + = r r e Part b) % 1859 . 7 071859 . 1 ) 4 / 07 . 1 ( 4 = =- + = eff r 28 . 644 , 151 \$ ) 4 / 07 . 1 ( 000 , 100 24 = + = FV Part c) =- + + + = 1 ) 1 ( ) 07 . 1 ( ) 1 ( 5 . 2 2 3 3 4 , 2 S S f 0.08 Part d) . 5 ) 450 ( 90 1 1 =- - =- = λ d dP P D m Part e) The bond price will decrease by a percent equal to the modified duration 854 . 4 ) 2 / 06 . 1 ( 5 ) 2 / 1 ( = + = + = λ D D m . Question 2 Refer to the following cash flow diagram: 4 \$ ) 6 (\$ 120 60 120 = + = AI The internal rate of return using continuous compounding is rt e C F AI P- + = + ) ( or ) 360 / 60 ( ) 106 ( 4 101 r e- = + solving for r gives % 687 . 5 106 105 ln ) 6 ( = - = r \$100 \$6 \$6 120 days 60 days \$101 AI Question 3 Part a) ( 29 18 . 98 \$ 1 . 1 108 yield the to equal is coupon the as 100 \$ 13 . 75 \$ 1 . 1 100 3 = = = = + = C B A P P P Part b) 909 ....
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This note was uploaded on 11/15/2008 for the course MS&E 242 taught by Professor Primbs during the Fall '06 term at Stanford.

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242MidTermSol2005 - Question 1 Part a 0418 10 12 1 12 10 =...

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